What is the integral of sin^3(4x)cos^3(4x) ?

The integral of sin^3(4x)cos^3(4x) is 1/4 (-(cos^4(4x))/4+(cos^6(4x))/6)+C

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integral of sin^3(4x)cos^3(4x)

\int sin^{3} (4 x) cos^{3} (4 x) d x

\frac{1}{4} (- \frac{cos ^{4} ( 4 x )}{4} + \frac{cos ^{6} ( 4 x )}{6}) + C

Solution steps

\int sin^{3} (4 x) cos^{3} (4 x) d x

Apply u-substitution

: \int sin^{3} (u) cos^{3} (u) \frac{1}{4} d u

= \int sin^{3} (u) cos^{3} (u) \frac{1}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int sin^{3} (u) cos^{3} (u) d u

Rewrite using trig identities

= \frac{1}{4} \cdot \int (1 - cos^{2} (u)) sin (u) cos^{3} (u) d u

Apply u-substitution

: \int - v^{3} (1 - v^{2}) d v

= \frac{1}{4} \cdot \int - v^{3} (1 - v^{2}) d v

Expand

- v^{3} (1 - v^{2}) : - v^{3} + v^{5}

= \frac{1}{4} \cdot \int - v^{3} + v^{5} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (- \int v^{3} d v + \int v^{5} d v)

\int v^{3} d v = \frac{v ^{4}}{4}

\int v^{5} d v = \frac{v ^{6}}{6}

= \frac{1}{4} (- \frac{v ^{4}}{4} + \frac{v ^{6}}{6})

Substitute back

= \frac{1}{4} (- \frac{cos ^{4} ( 4 x )}{4} + \frac{cos ^{6} ( 4 x )}{6})

Add a constant to the solution

= \frac{1}{4} (- \frac{cos ^{4} ( 4 x )}{4} + \frac{cos ^{6} ( 4 x )}{6}) + C

y^{^{'}} = (2 x + y)^{2}

y^{^{''}} + 7 y + 10 = 0, y (0) = 3, y^{^{'}} (0) = 18

\frac{d}{d s} (\frac{6 s ^{3} + 28 s ^{2} + 38 s + 15}{( s + 1 ) ^{2}})

\frac{d}{d x} (2 x^{3} - 9 x^{2} + 12 x - 3)

f (x) = \frac{1}{x ^{8}}

What is the integral of sin^3(4x)cos^3(4x) ?
The integral of sin^3(4x)cos^3(4x) is 1/4 (-(cos^4(4x))/4+(cos^6(4x))/6)+C